Can someone please tell me the answers and explain how you know?
Determine the qualities of the given sets. (Select all that apply.)
1) {(x, y)| x > 0, y > 0}
2) {(x, y)| 4 < x2 + y2 < 9}
3) {(x, y)| x2 + y2 ≤ 4 or 16 ≤ x2 + y2 ≤ 25 }
For each set decide if it is open, connected, simply-connected or none of the above. (choose all that apply)
Determine the qualities of the given sets. (Select all that apply.)
1) {(x, y)| x > 0, y > 0}
2) {(x, y)| 4 < x2 + y2 < 9}
3) {(x, y)| x2 + y2 ≤ 4 or 16 ≤ x2 + y2 ≤ 25 }
For each set decide if it is open, connected, simply-connected or none of the above. (choose all that apply)
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I'll show (1) and you can try the other two.
Let A = {(x,y) | x>0, y>0}. Notice this is the first quadrant in the xy-plane
Consider the functions f:R^2-->R defined by f(x,y)=x and g:R^2-->R defined by g(x,y)=y. Both f and g are continuous, and you can check that A = f^(-1)((0,infinity)) n g^(-1)((0,infinity)), where "n" means intersection. Now, both f^(-1)((0,infinity)) and g^(-1)((0,infinity)) are continuous preimages of open sets, so they are open. Hence, A is also open.
Visually, it's easy to see A is connected. In fact, A is path-connected because if (x,y) and (a,b) are two points in A, then the straight line segment connecting them lies entirely in A. To see this, notice that any point on that line segment can be represented as (tx+(1-t)a, ty+(1-t)b) for some value of t between 0 and 1. And the fact that x>0 and a>0 shows tx+(1-t)a>0 also. Similarly, ty+(1-t)b>0. Hence, (tx+(1-t)a, ty+(1-t)b) is in A. (Actually, we showed A is a "convex" set, which implies path-connected, which implies connected.)
Again, it is visually easy to see that A is simply-connected because it doesn't have any "holes." (On the other hand, the set in part (2) has a hole in the middle, so it is not simply-connected.) It's a little trickier to write out a more precise explanation than that though.
Let A = {(x,y) | x>0, y>0}. Notice this is the first quadrant in the xy-plane
Consider the functions f:R^2-->R defined by f(x,y)=x and g:R^2-->R defined by g(x,y)=y. Both f and g are continuous, and you can check that A = f^(-1)((0,infinity)) n g^(-1)((0,infinity)), where "n" means intersection. Now, both f^(-1)((0,infinity)) and g^(-1)((0,infinity)) are continuous preimages of open sets, so they are open. Hence, A is also open.
Visually, it's easy to see A is connected. In fact, A is path-connected because if (x,y) and (a,b) are two points in A, then the straight line segment connecting them lies entirely in A. To see this, notice that any point on that line segment can be represented as (tx+(1-t)a, ty+(1-t)b) for some value of t between 0 and 1. And the fact that x>0 and a>0 shows tx+(1-t)a>0 also. Similarly, ty+(1-t)b>0. Hence, (tx+(1-t)a, ty+(1-t)b) is in A. (Actually, we showed A is a "convex" set, which implies path-connected, which implies connected.)
Again, it is visually easy to see that A is simply-connected because it doesn't have any "holes." (On the other hand, the set in part (2) has a hole in the middle, so it is not simply-connected.) It's a little trickier to write out a more precise explanation than that though.